similar to: Trouble with Functions

hello, I am using following command classify_polarity(documents,algorithm="bayes",verbose=TRUE) output is: [1] "DOCUMENT 1" [1] "WORD: excited CAT: positive POL: strongsubj SCORE: 8.44419229853175" [1] "WORD: happy CAT: positive POL: strongsubj SCORE: 8.44419229853175" [1] "WORD: optimistic CAT: positive POL: weaksubj SCORE: 7.7510451179718" [1]

Hi All, I am trying to perform sentiment analysis using R with the help of the library(sentiment). I am using the function classify_emotions(sentiment). It is basically selecting a particular word from a sentence and with the help of the pr-defined score of that particular word, it is giving the sentiment score of the whole sentence. For example,in the sentence "I am happy",

Hi, I have been using R for a few months and I have this working code. Don't seen any problem but this takes a long time. So if I have about 30000 rows it takes a few minutes. If I have 100000 it does not seem to complete. Original Data: Proto Recv-Q Send-Q Local-Address Foreign-Address State tcp 0 0 172.20.100.2:60255

Ivan and Bert, thank you so much for your help. Ivan, your solution worked perfectly. I didn't really understand how to do string processing on a vector of strings, and your solution demonstrated it for me. I modified it to work with the tidyverses' stringr library in this way: bg3_race_sum <- bg3_race %>% left_join(pl_vars, by=c("variable" = "name"))

Emilio Ahora cuando quiero instalar los paquetes pdftools, magick y otros más me salen el siguiente error WARNING: Rtools is required to build R packages but is not currently installed. Please download and install the appropriate version of Rtools before proceeding: https://cran.rstudio.com/bin/windows/Rtools/ Installing package into ?C:/Users/bdominguez/Documents/R/win-library/3.6? (as ?lib?

Hello again: Here is a solution to the dates without leading zeros: pou1 <- function(x) { #Note: x is a data frame #Assume that Column 1 has the date #Column 2 has station #Column 3 has min #Column 4 has max library(stringr) w <- character(length=nrow(x)) z <- str_split(x[,1],"/") for(i in 1:nrow(x)) { u <-

You can use na.strings="" in read.table() or read.csv() library(stringr) vec1<-unlist(str_split(readLines(textConnection("3,7,11,,12,14,15,,17,18,19")),",")) ?vec1[vec1==""]<- NA ?vec1 # [1] "3"? "7"? "11" NA?? "12" "14" "15" NA?? "17" "18" "19" #or

Hi, May be this helps: vec<- "this is a nice text with nice characters" library(stringr) ?vec2<-unlist(str_match_all(vec,"\\w+")) #or # vec2<-str_split(vec," ")[[1]] res<-unique(lapply(vec2,function(x) which(!is.na(match(vec2,x))))) ?names(res)<- unique(vec2) res #$this #[1] 1 # #$is #[1] 2 # #$a #[1] 3 # #$nice #[1] 4 7 # #$text #[1] 5 # #$with #[1]

Buenas tarde a todo en s: Tenia la versión de R 3.6 y utilizaba la paquetería de pdftools para extraer información de archivos en pdf actualice la versión 3.6.1 y ya no reconoce la paquetería alguien que me pueda ayudar. Prácticamente no reconoce las funciones de pdftools library(pdftools) library(stringr)? library(NLP)? library(tm)? library(tesseract)? library(magick)?

Hi Paul, No need to collapse the information into a single text string, gregexpr() can take a vector of strings (sentences in your case). You can split your sentences up, number them how you want, then search for your pattern either via regex or via these extra packages you use which probably use the PCRE regex library anyway. However, as this is basically what you did, I'm not sure why

I have a problem to plot label (Year) only for significant values (in this case spoz and sneg). I use this code, but don't work with labels. library(ggplot2) ggplot(data1, aes(x = Year, y = value,fill=type,width=1))+ geom_bar(stat="identity",position="identity")+ scale_y_continuous(breaks = round(seq(-100, 100, by = 10),10))+ theme_bw() Thank you! the data used is:

Hello list, I want to make a large rulebased algorithm, to provide decision support for drug prescriptions. I have defined the algorithm in a function, with a for loop and many if statements. The structure should be as follows: 1. Iterate over a list of drug names. For each drug: 2. Get some drug related data (external dataset). Row of a dataframe. 3. Check if adaptions should be made to

Hello all! I have a problem to plot label (Year) only for significant values (in this case spoz and sneg). I use this code, but don't work with labels. library(ggplot2) ggplot(data1, aes(x = Year, y = value,fill=type,width=1))+ geom_bar(stat="identity",position="identity")+ scale_y_continuous(breaks = round(seq(-100, 100, by = 10),10))+ theme_bw() the data used is:

Dear R users, I have a problem to plot label (Year) only for significant values (in this case spoz and sneg). I use this code, but don't work with labels. library(ggplot2) ggplot(data1, aes(x = Year, y = value,fill=type,width=1))+ geom_bar(stat="identity",position="identity")+ scale_y_continuous(breaks = round(seq(-100, 100, by = 10),10))+ theme_bw() the data used

Hello, i have ported the BCP (Bride Control Protocol) patch for pppd 2.4.1 mentioned in http://lists.osdl.org/pipermail/bridge/2004-September/000619.html to pppd 2.4.2. The kernel patch still works without problems with kernel 2.4.30. Perhaps someone else could use this patch ... Here again some documentation i have found about the BCP patch somewere else: When pppd negotiates BCP, it tells

Hello all! I want to make a barplot with ggplot2. I want to view in the same chart the semn values (significant values (pointer over 50)). I try this code, but only for pointer values. ggplot(data, aes(x = Year, y = pointer)) + geom_bar(stat="identity") please help me with this problem. I use this data: Year variable pointer variable semn 1 1901 neg 0.00 sneg NA 2 1902 neg 0.00 sneg

Dear R users, There is a error message when I run the following code. It is used to load microarray data and use the top 1000 genes for training svm to classify test set . > library(e1071) Loading required package: class > f=read.table("F:\\lab\\ microarray analysis\\VEH LPS\\exprs.txt",

Hi all (really probably just Deepayan): In the plot below I want to add text on either side of each violin plot that indicates the number of observations that are either positive or negative. I'm trying to do this with ltext() and I've also monkeyed about with panel.text(). The code below is generally what I want but my calls to ltext() are wrong and I'm not sure how to fix them.

Dear community I am using ggraph to plot a network analysis. See part 2 in the working example. Besides different colors for different groups of nodes: --> geom_node_point(aes(size = V(network)$hub_score*200, color= as.factor(V(network)$community))) I additionally want to consider different colors for different edge groups The grouping is defined in the edge_list$relationship: negative

Dear r-help, I'm interested in finding a better way to add a column to a data frame when calculating the new column requires more than one conditional. For example, if I wanted to associate a character string in {"Pos","Neg","Zero"} with each number in the following data frame: > d <- data.frame(num = -2:2) > d num 1 -2 2 -1 3 0 4 1 5 2 I